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Bolt falls off rocket, find rocket acceleration

  1. Apr 10, 2005 #1
    A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 6.70 s later.

    Find the rockets acceleration.

    I dont understand how to set up this problem, ive never done something like this , asked some friends and they didnt know how to do it. Any help would be greatly appreciated :smile:
  2. jcsd
  3. Apr 10, 2005 #2
    Work backwards kind of. If it took 6.7 for the bolt to hit the ground, how high was it?

    [tex]d = v_{i}t + \frac{1}{2}at^2[/tex]

    vi = 0
    a = -32 ft\sec^2 or -9.8 m\sec^2
    t = 6.7 s

    Solve for d.

    Now, use the same equation and solve for "a" of the rocket.

    [tex]d = v_{i}t + \frac{1}{2}at^2[/tex]

    d = part a
    vi = 0
    t = 4

  4. Apr 10, 2005 #3

    Doc Al

    User Avatar

    Staff: Mentor

    Do it in two parts, but like this:
    Part 1: Find the height and speed of the rocket after 4 seconds in terms of the rocket acceleration. Use:
    [itex]h = 1/2 a t^2[/itex] and
    [itex]v = a t[/itex]

    Part 2: Realize that the initial height and speed of the bolt equals the height and speed of the rocket after 4 seconds (which was calculated in part 1). The height of the bolt after it leaves the rocket is:
    [itex]h = h_0 + v_0t -(g/2)t^2[/itex],
    using [itex]h_0[/itex] and [itex]v_0[/itex] (in terms of a) from part 1; set h = 0 when t = 6.7 seconds. Solve for a.
  5. Apr 10, 2005 #4
    thank you guys:cool:
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